Sry i was little wrong: nC0*(1/n)^n + nC2 *2*(1/n)^n + nC4*2^2*(1/n)^n+....+nCn*2^(n/2)*(1/n)^n when n is even
nC0*(1/n)^n + nC2 *2*(1/n)^n + nC4*2^2*(1/n)^n+....+nCn-1*2^(n-1/2)*(1/n)^n when n is odd On Mon, Sep 12, 2011 at 10:08 PM, Piyush Grover <[email protected]>wrote: > it should be: > > (1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n/2)+1 terms) = {2^(1 + n/2) - > 1}*(1/n)^n when n even > > (1/n)^n * (1 + 2 + 2^2 + 2^3 +....(n-1/2)+1 terms) = {2^(1 + (n-1)/2) - > 1}*(1/n)^n when n is odd > > > -Piyush > > > On Mon, Sep 12, 2011 at 8:01 PM, sandeep nagamalli > <[email protected]>wrote: > >> I think it is 1 / (2N) >> >> (1/N) * (1/N)*(N/2) = 1/(2N) >> >> >> On Mon, Sep 12, 2011 at 6:33 PM, Akash Mukherjee <[email protected]>wrote: >> >>> this is a case, but isnt der a case of circular permutation 2?? >>> what say abt dis >>> 0,1 >>> 2,3 >>> 4,5 >>> 0,1 >>> 2,3 >>> 4,5 >>> >>> it wrks i gues?? >>> >>> >>> On Mon, Sep 12, 2011 at 12:56 PM, teja bala >>> <[email protected]>wrote: >>> >>>> >>>> >>>> I think it is 1/N >>>> >>>> let N=6 that means rand(6)= takes values 0-5 i.e 6 values. >>>> rand(m)=6 values >>>> rand(n)=6 values >>>> total combinations 6*6=36 values set but among dem array will change >>>> only for >>>> (0,0)(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)=6values of N >>>> >>>> So 6/36=1/6 >>>> >>>> If we generalize it 1/N >>>> >>>> correct me if i'm wrong? >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To post to this group, send email to [email protected]. >>>> To unsubscribe from this group, send email to >>>> [email protected]. >>>> For more options, visit this group at >>>> http://groups.google.com/group/algogeeks?hl=en. >>>> >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> >> >> -- >> Thanks&Regards: >> -sandeep >> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
